Numbers form an integral part of our lives. In this chapter, we will learn about the different types of numbers and the different categories under which they fall. The concepts discussed here will be your first step towards understanding the mathematics requirements to clear MBA entrance exams. As we proceed with this chapter, you will realize that you have already learned many of the concepts in school. This would further help build confidence in you.
A number is a mathematical concept used for counting, measuring, or labelling things. Numbers play a key role in arithmetic calculations.
Real numbers include the set of rational and irrational numbers.
Rational Numbers
Irrational Numbers
The set including all whole numbers and their negatives is called a set of integers. It is denoted by Z, and Z = { ∞, …  3,  2,  1, 0, 1, 2, 3, ……. ∞}.
The set of numbers that includes all natural numbers along with zero is called whole numbers.
Whole numbers are also known as nonnegative integers.
The numbers 1, 2, 3, 4, 5…are known as natural numbers. The set of natural numbers is denoted by N.
All numbers divisible by 2 are called even numbers.
Example: 2, 4, 6, 8, 10.…
Even numbers can be expressed in the form 2n, where n is an integer. Thus 0,  2, − 6, etc. are also even numbers.
All numbers not divisible by 2 are called odd numbers.
Odd numbers can be expressed in the form (2n + 1) where n is an integer.
Example:
 1, − 3, − 5…
1, 3, 5, 7, 9…
A natural number that has no other factors besides itself and unity is a prime number.
Examples: 2, 3, 5, 7, 11, 13, 17, 19 …
On the contrary, a number which has more than two factors is called a composite number.
Important Observations About Prime Numbers
Some Properties of Prime Numbers
How to check whether a number is prime or not
To check whether a number N is prime, adopt the following process
To illustrate:
The value of √239 lies between 15 to 16. Hence, take the value of z as 16.
Prime numbers less than or equal to 16 are 2,3,5,7,11 and 13. 239 is not divisible by any of these. Hence, you can conclude 239 is a prime number.
Finding Prime Numbers: The Short Cut Method
Checking Whether a Number is Prime (For Numbers below 49)
Checking Whether a Number is Prime (For Numbers above 49 and below 121)
Checking Whether a Number is Prime (For Numbers above 121 and below 169)
Must Know
1 to 25 ⇒ 9 prime numbers
1 to 50 ⇒ 15 prime numbers
1 to 100 ⇒ 25 prime numbers
1 to 200 ⇒ 45 prime numbers
Example: If a, a + 2, a + 4 are consecutive prime numbers. Then how many solutions ‘a’ can have?
(a) One
(b) Two
(c) Three
(d) More than three
Correct Answer is Option (a)
 No even value of ‘a’ satisfies this. So ‘a’ should be odd.
 But out of three consecutive odd numbers, at least one number is a multiple of 3.
 So, the only possibility is a = 3 and the numbers are 3, 5, 7.
A composite number has other factors besides itself and unity.
Example: 8, 72, 39, etc.
Note: 1 is neither a prime number nor a composite number.
A number is said to be a perfect number if the sum of all its factors excluding itself (but including 1) is equal to the number itself or the sum of all the possible factors of the number is equal to twice the number.
A fraction denotes the part or parts of a unit.
Several types of fractions are:
A decimal number is a number with a decimal point in it. For example, 1.5, 3.22, 829.234
Converting recurring fraction into decimals:
All recurring decimals can be converted into fractions. Some of the common types can be 0.33…..,0.1232323…, 5.33…., 14.23636363…. etc.
(a) Pure Recurring to Fractions
Funda 1: If a number is of the form of 0. ababab……. then divide the repeating digits with as many 9’s as we have repeated digits.
Example: 0.363636...= 36/99 = 4/11
(b) Mixed Recurring to Fractions
Funda 2: If N = 0. abcbcbc…. Then N = abc  a/990 = Repeated & non Repeated digits  Non Repeated digits/ As many 9's as repeated digits followed by as many zero as nonrepeated digits
Example: 0.25757...= 257  2/990 = 255/990 = 17/66
Funda 3. If N = a. bcbc…. Then
Write N = a + 0. bcbc….Proceed as Funda 1
5.3636… = 5 + 0.3636… = 5 + 36/99 = 59/11
A number line is a straight line that extends infinitely in both directions, with negative infinity on the left and positive infinity on the right.
Number Line
Note: Even when we divide a negative number by a natural number N, the remainder remains nonnegative. For instance, when we divide –32 by 7, the remainder is actually +3, not –4, which might be surprising. This is because the remainder is always nonnegative. So, when we divide –32 by 7, the quotient is –5 and the remainder is +3.
Theorems of Divisibility
1. Divisibility Rule of 1
2. Divisibility Rule of 2
3. Divisibility Rules for 3
4. Divisibility Rule of 4
5. Divisibility Rule of 5
6. Divisibility Rule of 6
7. Divisibility Rules for 7
The rule for divisibility by 7 is a bit complicated which can be understood by the steps given below:
Example: Is 1073 divisible by 7?
 From the rule stated remove 3 from the number and double it, which becomes 6.
 Remaining number becomes 107, so 1076 = 101.
 Repeating the process one more time, we have 1 x 2 = 2.
 Remaining number 10 – 2 = 8.
 As 8 is not divisible by 7, hence the number 1073 is not divisible by 7.
8. Divisibility Rule of 8
9. Divisibility Rule of 9
10. Divisibility Rule of 10
11. Divisibility Rules for 11
If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.
In order to check whether a number like 2143 is divisible by 11, below is the following procedure:
12. Divisibility Rule of 12
13. Divisibility Rules for 13
Example 1. N = (18n^{2} + 9n + 8)/n; where N belongs to integer. How many integral solutions of N are possible?
The given expression can be broken as: 18n^{2}/n + 9n/n + 8/n. This gives us: 18n + 9 + 8/n.
 Now we can see that whatever the value of ‘n’, 18n + 9 will always give an integral value. Therefore, it now depends upon 8/n the only ⇒ n can have any integral value, which is a factor of 8.
 The integers, which will satisfy this condition are ±1, ±2, ±8, ±4. Thus, in total, n can take 8 values.
Example 2. N = 960. Find the total number of factors of N.
 The number of divisors of a composite number: If D is a composite number in form D = a^{p} × b^{q} × c^{r}, where a, b, c are primes, then the no. of divisors of D, represented by n is given by n = (p+1)(q+1)(r +1).
 Following the same, after dividing 960 into prime factors: 2^{6} × 3^{1} × 5^{1}, we can calculate the total number of factors as (6+1). (1+1).(1+1) = 28.
Example 3. Find the unit digit of the following expression: (123)^{34} × (876)^{456} × (45)^{86}.
 Whenever an even unit digit Number is present and a Number with 5 at the unit digit are present, they will always give a 0 at the unit digit, no matter if any other number is present or not.
Example 4. What will be the number of zeroes at the end of the product of the first 100 natural numbers?
Example 5. Which letter should replace the $ in the number 2347$98, so that it becomes a multiple of 9?
 As you know that if the sum of all the digits is divisible by 9, then the number is divisible by 9.
 Now sum of the given digits is 2 + 3 + 4 + 7 + 9 + 8 = 33 + $. Now think the next multiple of 9 after 33 i.e. 36. This means you add 3 in this. The value of $ is 3.
Example 6. In a party, there are 20 people present. If each of them shakes hands with all the other persons, in total, how many handshakes will take place?
 Out of 20 persons, the first person will shake hand with 19 persons.
 The second person will shake hand with 18 persons (because he has already shaken hand with first person).
 The third person will shake hand with 17 persons and so on.
 The second last person shakes hand with only one person.
 And last will shake hand with none (because he has already shaken hand with all persons).
 In order to find the total number of handshakes, you have to add all the natural numbers from 1 to 19, i.e. ∑ 19. ∑19 = 19 x 20/2 = 190 handshakes.
Example 7. What is the remainder when 2354789341 is divided by 11?
 Odd place digit sum (O) = 1 + 3 + 8 + 4 + 3 = 19.
 Even place digits sum (E) = 4 + 9 + 7 + 5 + 2 = 27.
 Difference (D) = 19  27 =  8
 Remainder = 11  8 = 3.
Tip: When any number with even number of digits is added to its reverse, the sum is always divisible by 11. e.g. 2341 + 1432 = 3773, which is divisible by 11.
⇒ Any number written 6 times consecutively will be divisible by 7 and 13.
Example 8. If 567P55Q is divisible by 88; Find the value of P + Q.
(a) 11
(b) 12
(c) 5
(d) 6
(e) 10
Correct Answer is Option (e)
 The number is divisible by 8 means; the number formed by the last 3 digits should be divisible by 8, which are 55 Q. Only Q = 2 satisfies this.
 From the divisibility rule of 11, (2 + 5 + 7 + 5)  (5 + P + 6) is divisible by 11. So 8P is divisible by 11.
 If P= 8, then only it is possible. So P = 8 and Q = 2.
 So, P + Q = 10.
Example 9. If the first 100 natural numbers are written side by side to form a big number and it is divided by 8. What will be the remainder?
(a) 1
(b) 2
(c) 4
(d) 7
(e) Cannot be determined
Correct Answer is Option (c)
 The number is 1234…..9899100.
 According to the divisibility rule of 8, we will check only the last 3 digits.
 If 100 is divided by 8, the remainder is 4.
Example 10. What will be the remainder when 4444……..44 times is divided by 7?
(a) 1
(b) 2
(c) 5
(d) 6
(e) 0
Correct Answer is Option (b)
 If 4 is divided by 7, the remainder is 4.
 If 44 is divided by 7, the remainder is 2.
 If 444 is divided by 7, the remainder is 3.
 By checking like this, we come to know that 444444 is exactly divisible by 7.
 So if we take six 4’s, it is exactly divisible by 7.
 Similarly, twelve 4’s is also exactly divisible by 7 and 42 4’s will be exactly divisible by 7. So out of 44, the remaining two 4,s will give a remainder of 2.
Two or more numbers that share no common factor are called coprime or relatively prime. In other words, their greatest common factor is 1.
If two numbers m and n are relatively prime, and a natural number x is divisible by both m and n individually, then x is also divisible by mn.$mn$
Key Concept 1: Recognizing two numbers as coprime is important when they appear in the denominators of fractions.
Example: Find all fourdigit numbers of the form 25x7y that are divisible by 36.
Sol: Since 36 is the product of two coprime numbers, 4 and 9, the number 25x7y must be divisible by both 4 and 9 for it to be divisible by 36.
Divisibility by 4: For divisibility by 4, the last two digits of the number (7y) must be divisible by 4. The possible values for y that make 7y divisible by 4 are 2 and 6. So, y can be 2 or 6.
Divisibility by 9: For divisibility by 9, the sum of the digits of the number must be divisible by 9. The sum of the digits is $\mathrm{2+5+x+7+y.}$
If y=2, the sum is 2+5+x+7+2=16 + x.
For this to be divisible by 9, 16 + x must be a multiple of 9. The possible value of x is 2 (since 16 + 2 = 18, which is divisible by 9).
If y=6, the sum is 2+5+x+7+6=20 + x.
For this to be divisible by 9, 20 + x must be a multiple of 9. The possible value for x is 7 (since 27 is divisible by 9).$x$Thus, the numbers 25272 and 25776 are divisible by 36.
Key Concept 2: Another way to determine if two numbers are coprime is by examining their prime factors. For two or more numbers to be coprime, their prime factors must not have any common elements.
For example, if A and B are coprime, and A = 2^{n}x 3^{m}, then the prime factors of B must not include 2 or 3. Instead, would be made up of other prime numbers like 5, 7, 11, etc., but it cannot have 2 or 3 as factors.
Example 1. If ‘X’ is an even number; Y is an odd number, then which of the following is even?
(a) X^{2} + Y
(b) X + Y^{2}
(c) X^{2} + Y^{2}
(d) X^{2}Y^{2}
(e) None of these
Correct Answer is Option (d)
 Since X is even, X^{2} is even.
 Y is odd, Y^{2} is odd.
 So options (1), (2), (3) are even + odd = odd.
 Option (4) is (even) (odd) = Even.
Example 2. What is the difference between 0.343434....…and 0.2343434…… in fraction form?
(a) 6/55
(b) 6/11
(c) 9/55
(d) 9/13
(e) 5/11
Correct Answer is Option (a)
0.343434.....= 34/99 and 0.23434.... = 234  2/990 = 232/990
∴ Difference = 34/99  232/990 = 108/990 = 6/55
Example 3. How many of the following numbers are divisible by at least 3 distinct prime numbers 231, 750, 288 and 1372?
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4
Correct Answer is Option (c)
 231 = 3 × 7 × 11 (3 prime factors)
 750 = 2 × 3 × 53 (3 prime factors)
 288 = 25 × 32 (only 2 prime factors)
 1372 = 22 × 73 (only 2 prime factors)
So, only 231 & 750 has 3 prime factors
Example 4. n^{3} + 6n^{2} + 11n + 6 (where n is a whole no) is always divisible by
(a) 4
(b) 5
(c) 6
(d) 8
(e) 12
Correct Answer is Option (c)
 n^{3} + 6n^{2} + 11n + 6 = (n + 1) (n + 2) (n + 3).
 Product of 3 consecutive numbers is always divisible by 3! = 6. (or)Take n = 0, 1, 2, 3 and check it is always divisible by 6.
Example 5. What is the remainder, if 351 × 352 × 353 ×        × 356 is divided by 360?
(a) 0
(b) 1
(c) 2
(d) 3
(e) 359
Correct Answer is Option (a)
 Since the given is the product of 6 consecutive numbers, it is always divisible by 6! = 720.
 It is divisible by 360 also. So, the remainder will be 0.
Let us learn the process of finding HCF with the help of some solved examples.
Q: Find the HCF of 96, 36 and 18.
Sol: First break each number into their prime factors.
96 =2 x 3 x 2 x 2 x 2 × 2
36 = 2 x 3 x 2 x 3
18 = 2 x 3 x 3
Take all the common factors of these numbers and multiply them.
Therefore, the HCF of 96, 36 and 18 is the product of the highest number of common factors in the given numbers i.e., 2 x 3 = 6. In other words, 6 is the largest possible integer, which can divide 96, 36 and 18 without leaving any remainder.
Q: Find the HCF of 42 and 70.
Sol:
42 = 3 x 2 x 7
70 = 5 x 2 x 7
Hence, HCF is x 7 = 14.
Q: Find the HCF of numbers 144, 630 and 756
Sol:
144 = 2^{4} x 3^{2}
630 =2×3^{2}x 5 x7
756 = 2^{2} x 3^{3} x7
Hence, HCF of 144, 630, 756 = 2 x 3^{2} = 18.
Important:
Let's learn a faster way to find the HCF (Highest Common Factor) of a set of numbers using examples.
Example 1: Find the HCF of 39, 78, and 195.
Sol: Look at the stepbystep method:
Look at the differences between the numbers.
 Take the difference between any two numbers. The HCF must be a factor of these differences.
 The differences are:
 $7839=39$78−39=39
 $19539=156$195−39=156
 $19578=117$195−78=117
 So, the HCF must divide 39, 156, and 117.
Start with the smallest difference (39).
 The HCF must be a factor of 39. The factors of 39 are:
 1, 3, 13, and 39.
Check if the largest factor (39) divides all the numbers.
 Does 39 divide 78? Yes.
 Does 39 divide 195? Yes.
 So, the HCF of 39, 78, and 195 is 39.
Example 2: Find the HCF of 39, 78, and 182.
Sol:
 Find the differences between the numbers.
 $7839=3978\; \; 39\; =\; 39$78−39=39
 $18239=143182\; \; 39\; =\; 143$182−39=143
 $18278=104182\; \; 78\; =\; 104$182−78=104
 Check the factors of 39 (1, 3, 13, 39).
 Does 39 divide 182? No.
 Try the next largest factor, 13.
So, the HCF of 39, 78, and 182 is 13.
 Does 13 divide 39? Yes.
 Does 13 divide 182? Yes.
Why Does This Work?
The Least Common Multiple of two or more numbers is the smallest number which is exactly divisible by all of them. In other words, it is the product of the highest powers of all the prime factors of the given numbers.
To find the LCM of given numbers:
Let us take some solved examples.
Question: Find the LCM of 96, 36 and 18.
Solution:
First, Break each number into their prime factors
96 = 2 x 2 x 2 x 2 x 2 x 3 =2^{5}x 3^{1}
36 = 2 x 2 x 3 x 3 = 2^{2} x 3^{2}
18 = 2x 3 x 3= 2^{1} x 3^{2}
Therefore, LCM of 96, 36 and 18 is the product of the highest powers of all the prime factors, i.e. 2^{5} x 3^{2}= 32 x 9 = 288.
That is, 288 is the smallest integer, which is divisible by 96, 36 and 18 without leaving any remainder.
Question: Find the LCM of 42 and 70
Solution:
42 = 3 x 2 x 7
70 = 5 x 2 x 7
Hence, LCM is 2 x 3 x 5 x 7 = 210.
Apart from the method of prime factorization, there is another method of finding the LCM of given numbers and the method is known as the long division method. This method is quite helpful in getting LCM quickly if there are three or more than three numbers.
Write the numbers, separated by commas. Then divide them by prime factors in ascending order (e.g., 2, 3, 5, 7, etc.) one at a time. Then, after each division, write ‘ the quotient of each number that gets completely divided by the divisor (the prime number) below it. Leave the undivided numbers as they are. Continue doing this till you get prime factors as quotients in each column, The product of all the prime factors (divisors and quotients) will be the LCM.
Question: Find the LCM of 8, 12, 15 and 21
Solution:
Hence, LCM is 2x 2x 2x3x 5x 7= 840.
Important:
1. HCF of A, B and C is the highest divisor which can exactly divide A, B, and C.
2. LCM of A, B and C is the lowest dividend which is exactly divisible by A, B, and C.
There is one very important relationship, given below, between two numbers and their HCF and LCM. Many problems have appeared in
various competitive exams based on this relationship.
Example: LCM and HCF of two numbers are 2079 and 27 respectively. If one of the numbers is 189, find the other number.
Sol: The other number will be=
Hence, the required number =
Example: Two numbers are in the ratio 3: 5 and their LCM is 1500. Find the HCF of the numbers.
Sol:
Let the two number be 3X and 5X.
Hence their HCF = X using the formula given above, we get LCM = 3×5×X=15X
Or, 15X=1500 => X=100.
Therefore, the HCF of the numbers is 100
We can use the following direct formula.
HCF of fractions =
Similarly, LCM of fractions =
Example: Find the HCF and LCM of the two fractions 6/9, 12/15.
Sol:
The numerators are 6 and 12, their HCF = 6 and LCM =12
The denominators are 9 and 15, their LCM = 45 and HCF=3
Hence, the required HCF = and the LCM = 12/3=4.
Example: What is the greatest possible rate at which a man can walk 51 km and 85 km in an exact number of minutes?
Sol: To find the greatest possible rate at which the man can walk 51 km and 85 km in an exact number of minutes, we need to determine the largest common factor of 51 and 85 in terms of kilometers per minute. This problem is asking for the HCF (Highest Common Factor) of the distances 51 km and 85 km.
StepbyStep Process:
1. Find the HCF of 51 and 85:
 The prime factors of 51 are: 51=3× 17
 The prime factors of 85 are: 85=5×17
The common factor between 51 and 85 is 17.
2. Result: The greatest possible rate at which the man can walk in an exact number of minutes is 17 km per minute, since 17 is the largest number that divides both 51 and 85 exactly.
Thus, the greatest possible rate is 17 kilometers per minute.
Important points on the remainder
E.g., 5 divided by 6, the remainder is 5 only.
E.g., Remainder of when 4 is divided by 6 is 4 and NOT 2.
The Remainder Theorem helps us find the remainder when a product is divided by a number without having to calculate the whole product. Let's break it down with the example of 17 x 23 divided by 12.
Rewrite the numbers: We can express the numbers 17 and 23 in a way that includes 12:
17 = 12 + 5
23 = 12 + 11
So, we can rewrite the expression:
17 x 23 = (12+5) x (12+11)Expand the expression: Using the distributive property, expand the expression
(12+5)(12+11) = 12 x 12 + 12 x 11 + 5 x 12 + 5 x 11Identify the terms:
12 x 12 (is divisible by 12, so it leaves a remainder of 0)
12 x 11 (divisible by 12)
5 x 12 (divisible by 12)
Last term: 5 x 11Focus on the last term: The only term that affects the remainder when divided by 12 is 5 x 11
5 x 11 = 55Find the Remainder: Now, we need to find the remainder when 55 is divided by 12:
55 / 12 = 4 with a remainder of 7Thus, the remainder when 17 x 23 is divided by 12 is 7.
The Remainder Theorem simplifies the process of finding remainders by focusing only on the relevant terms after rewriting the numbers in relation to the divisor. In our example, we only needed to consider the last term $\mathrm{5\times 11\; to\; find\; the\; remainder.}$
Let us take an example to understand it:
We know that remainder obtained when 15 is divided by 4 is 3 as 15 is 3 more than the nearest multiple of 4 i.e. 12.
Here we are comparing 15 with 12 (the largest multiple of 4 less than equal to 15 ) and found that 15 is 3 more than 12 and hence the excess portion is 3 which is the remainder.
Instead of comparing with 12, if we compare 15 with 16 (another multiple of 4 close to 15) we can say that 15 is 1 LESS than 16. Or there is a deficiency of 1 in 15 to make it divisible by 4. This deficient number is called a negative remainder. i.e. we can say that the remainder when 15 is divided by 4 is 1.
Conversion from negative remainder to the corresponding positive remainder and vice versa
Note: Product, addition and subtraction of any two or more numbers has the same remainder when divided by any natural number, as the corresponding product, addition and subtraction of their remainders.
Let us understand this with an example:
Take a number N =. Now let us find the remainder when N is divided by 5.
One approach is that we first calculate the value of N =i.e. which is equal to 197. And then dividing this number by 5 to get the remainder = 2.
The other approach is that we calculate the remainders of each number given in N by dividing it by 5 and then use the corresponding mathematical operators with these remainders.
Remainder obtained when 24 is divided by 5 = 4
Remainder obtained when 8 is divided by 5 = 3
Remainder obtained when 12 is divided by 5 = 2
Remainder obtained when 7 is divided by 5 = 2
Replacing numbers with their corresponding remainders we get,Since 12 is greater than 5, we divided 12 by 5 again to get the final remainder = 2.
Note: The second approach might sound difficult or redundant for the given example. But the detailed explanations are given for the understanding purpose, we will see that how applying this second approach we can solve advanced problems.
Applying both positive remainder and the concept of negative remainder simultaneously in the above problem with the second approach:
Remainder obtained when 24 is divided by 5 = 1
Remainder obtained when 8 is divided by 5 = 2
Remainder obtained when 12 is divided by 5 = 2
Remainder obtained when 7 is divided by 5 = 2
Replacing numbers with their corresponding remainders we get,Hence the final remainder = 2.
Question: What is the remainder when 123 × 124 × 125 is divided by 9?
Solution:
Remainder obtained when 123 is divided by 9 = 3
Remainder obtained when 124 is divided by 9 = 2
Remainder obtained when 123 is divided by 9 = 1
Final remainder = (3)(2)(1) = 6. The required positive remainder = 96 = 3.
Question: What is the remainder when 1! + 2! + 3! + …. + 100! Is divided by 5.
Solution:
Observe that in the series 5! onwards every number is divisible by 5 i.e. the remainder in each case is 0.
So the required remainder is obtained by dividing only the first 4 numbers i.e.
When we observe the behavior of these digits, they all have the same unit's digit as the number itself when raised to any power, i.e. 0^{n} = 0, 1^{n} =1, 5^{n} = 5, 6^{n} = 6.
5 ^{2} = 25: Unit digit is 5, the number itself.
1^{6} = 1: Unit digit is 1, the number itself.
0^{4 }= 0: Unit digit is 0, the number itself.
6^{3} = 216: Unit digit is 6, the number itself.
Let's apply this concept to the following example.
Example: Find the unit digit of the following numbers:
Both these numbers have a cyclicity of only two different digits as their unit's digit.
4^{2} = 16: Unit digit is 6.
4^{3} = 64: Unit digit is 4.
4^{4} = 256: Unit digit is 6.
4^{5} = 1024: Unit digit is 4.
9^{2} = 81: Unit digit is 1.
9^{3} = 729: Unit digit is 9.
It can be observed that the unit digits 6 and 4 are repeating in an oddeven order. So, 4 has a cyclicity of 2. Similar is the case with 9.
It can be generalized as follows:
Example: Find the unit digit of the following numbers:
These numbers have a power cycle of 4 different numbers.
2^{1} = 2, 2^{2} = 4, 2^{3} = 8 & 2^{4} = 16 and after that it starts repeating.
So, the cyclicity of 2 has 4 different numbers 2, 4, 8, 6.
3^{1} = 3, 3^{2} = 9, 3^{3} = 27 & 3^{4} = 81 and after that it starts repeating.
So, the cyclicity of 3 has 4 different numbers 3, 9, 7, 1.
7 and 8 follow similar logic.
So these four digits i.e. 2, 3, 7 and 8 have a unit digit cyclicity of four steps.
The concepts discussed above are summarized in the given table.
The power concept or cyclicity of a number helps us figure out the last digit of a number raised to a large power without actually calculating the whole thing. It's based on a repeating pattern that depends on the last digit of the number. A table helps us predict this last digit. Also, digits that appear once or twice are repeated every four times. So, each digit repeats every four times.
Step 1
Step 2
Q. 12 × 15 × 5 × 24 × 13 × 17
(a) 0
(b) 1
(c) 2
(d) 3
Ans: (c)
Sol: 2^{2}∗3∗3∗5∗5∗2^{3}∗3∗13∗17 = so pairs of (5*2) are 2, so we have 2 zeroes.
Example: n! has 13 zeroes. The highest and least values of n are?
(a) 57 and 58
(b) 59 and 55
(c) 59 and 6
(d) 79 and 55
Ans: (b)
Sol: At 55 we get 13 zeroes, since we know that 50! is 12 zeroes so till 54! we will have 12 zeroes. So 55 to 59! will have 13 zeroes.
1. Decimal number system (Base 10)
2. Binary number system (Base 2)
3. Octal number system (Base8)
4. Hexadecimal number system (Base 16)
1. Decimal number system (Base 10)
The decimal number system, with a base of 10, employs digits 0 to 9. Positions to the left of the decimal point represent powers of 10, indicating units, tens, hundreds, etc.
Example: 3567.89
In this example, the digit 3 is in the thousands place (10^{3}), the digit 5 is in the hundreds place (10^{2}), the digit 6 is in the tens place (10^{1}), the digit 7 is in the units place (10^{0 or 1}), the digit 8 is in the tenths place and the digit 9 is in the hundredths place.
2. Binary number system (Base 2)
Binary, or base2, has only two digits, 0 and 1. Binary numbers, like 110101, consist of combinations of these two digits.
Example: 110101
The rightmost digit is 2^{0}, the next is 2^{1}, then 2^{2}, and so on. The binary number 110101 is equivalent to (1 * 2^{5}) + (1 * 2^{4}) + (0 * 2^{3}) + (1 * 2^{2}) + (0 * 2^{1}) + (1 * 2^{0}) = 53 in the decimal system.
3. Octal number system (Base8)
Octal, with a base of 8 and digits 0 to 7, is used in computing. Converting octal to decimal follows the same process as standard decimal conversion.
Example: 745
In example 745, the rightmost digit is 8^{0}, the next is 8^{1}, and the leftmost is 8^{2}. Converting this to decimal, the octal number 745 is equivalent to (7 * 8^{2}) + (4 * 8^{1}) + (5 * 8^{0}) = 485 in the decimal system.
4. Hexadecimal number system (Base 16)
Hexadecimal uses base 16 and represents numbers from 0 to 9, then employs A to F for values 10 to 15. It's a system commonly used in computing.
Example: 1A3F
In the example 1A3F, it is equivalent to (1 * 16^{3}) + (10 * 16^{2}) + (3 * 16^{1}) + (15 * 16^{0}) = 6719 in the decimal system.
Q1: Find the last digit of 5555^{2345} + 6666^{5678}
(a) 1
(b) 3
(c) 5
(d) 7
Ans: a
As the last digit depends only on last digits, consider the powers only for last digits, that is 5^{2345} + 6^{5678} .
As we know, any power of 5 ends only with 5 and any power of 6 ends only with 6.
The last digit of 52345 + 65678 = 5 + 6 = 11 = 1
Hence, option (A) is the correct answer.
Q2: If in a two digit number, the digit at the unit place is z and the digit at the tens place is 8, then the number is
(a) 80z + z
(b) 80 + z
(c) 8z + 8
(d) 80z + 1
Ans: (b)
Digit at unit’s place = z
The digit at ten’s place = 8
= 2digit number = (10×8)+(1×z)
= 80 + z
Q3: How many trailing zeroes (zeroes at the end of the number) does 60! have?
(a) 14
(b) 12
(c) 10
(d) 8
Ans: (a)
 To start with, the number of trailing zeroes in the decimal representation of a number = highest power of 10 that can divide the number.
For instance,
3600 = 36 * 102
45000 = 45 * 103 In order to approach this question, let us first see the smallest factorial that ends in a zero.
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
Now, 5! ends in a zero as we have get a product of 10 when we compute 1 * 2 * 3 * 4 * 5.
10 is 2 * 5, so we get a factor of 10 every time we get a 2 and a 5 in the factorial.
So, 5! has 1 zero. The factorial that ends with 2 zeroes is 10!
15! has 3 zeroes.
20! has 4 zeroes and so on.
An extra zero is created every time a 2 and 5 combine. Every even number gives a two, while every fifth number gives us a 5. Now, the critical point here is that since every even number contributes at least a 2 to the factorial, 2 occurs way more frequently than 5. So, in order to find the highest power of 10 that can divide a number, we need to count the highest power of 5 that can divide that number. We do not need to count the number of 2’s in the system as there will be more than 2’s than 5’s in any factorial.
 Now, every multiple of 5 will add a zero to the factorial. 1 * 2 * 3 *.......59 * 60 has twelve multiples of 5. So, it looks like 60! will end in 12 zeroes. But we need to make one more adjustment here.
 25 is 5^{2}, so 25 alone will contribute two 5’s, and therefore add two zeroes to the system. Likewise, any multiple of 25 will contribute an additional zero.
 So, 20! has 4 zeroes, 25! has 6 zeroes.
60! will have [60/5] zeroes arising due to the multiples of and an additional [60/25] due to the presence of 25 and 50.
{We retain only the integer component of [60/25] as the decimal part has no value} So, 60! will end with 12 + 2 zeros. = 14 zeros.
 In general, any n! will end with zeroes.
 Generalizing further, in case we want to find the highest power of 3 that divides n!, this is nothing but
 The highest power of 7 that divides n! is
 In case of a composite number, we need to break into the constituent primes and compute the highest power that divides the number.
 For instance, if we want to find the largest power of 15 that divides n!, this will be driven by the highest powers of 3 and 5 that divide n!. Similar to the scenario we saw with trailing zeroes, we can observe that there will definitely be at least as many 3’s than 5’s in any factorial. So, the highest power of 15 that divides n! is simply [n/5] + [n/25] + [n/125] + [n/625]............
 Hence the answer is "14"
Q4: The number of zeros at the end of the product of
(a) 42
(b) 53
(c) 1055
(d) None of these
Ans: (a)
The number of zeros at the end of 222^{111} × 35 ^{53} is 53.
The number of zeros at the end of (7!)^{6!}×(10!)^{5!} is 960.
The number of zeros at the end of 42^{42}×25^{25} is 42.
Thus the number of zeros at the end of the whole expression is 42.
Q5. How many even integers n, where 100 ≤ n ≤200 , are divisible neither by seven nor by nine
(a) 40
(b) 37
(c) 39
(d) 38
Ans: (c)
Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51  (7+61) = 39 There is one more method through which we can nd the answer. Since we have to nd even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.
Q6. How many pairs (a, b) of positive integers are there such that a ≤ b ab = 4^{2017} ?
(a) 2018
(b) 2019
(c) 2017
(d) 2020
Ans: (c)
ab = 4^{2017} = 2^{4034}
The total number of factors = 4035.
out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017.
And since the given number is a perfect square, we have one set of equal factors.
.'. many pairs (a, b) of positive integers are there such that a ≤ b and ab = 4^{2017}= 2018
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